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How to finish the proof of: If $N\unlhd G=H\times K$, then $N$ is abelian or $N$ intersects $H \times \{e\}$ or $\{e\} \times K$ non trivially

Suppose $N$ is normal, and nonabelian. Since it is nonabelian, there exists an element $(h,k)\in N$ such that either $h$ is not central in $H$, or $k$ is not central in $K$ (if both $h\in Z(H)$ and $k\...
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Let $H,K\le G$, abelian $G$, and let $\phi:G\to H$ be a hom. s.t. 1) $\phi(h)=h\forall h\in H$, 2) $\text{Ker }\phi=K$. Show $G=H\oplus K$

This is done in three steps: $H,K\unlhd G$ since each subgroup of an abelian group is normal. Let $g\in H\cap K$. Since $g\in H$, we have $\phi(g)=g$, but $g\in K$, so $0=\phi(g)=g$. Thus $H\cap K=\{...
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Let $H,K\le G$, abelian $G$, and let $\phi:G\to H$ be a hom. s.t. 1) $\phi(h)=h\forall h\in H$, 2) $\text{Ker }\phi=K$. Show $G=H\oplus K$

Hint The short exact sequence $$1\to K\hookrightarrow G\overset{\phi}{\to }H\to1$$ is left split. The morphism $\psi:G\to K$ being $\psi(x)=x-\phi(x)$.
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3 votes
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Let $H,K\le G$, abelian $G$, and let $\phi:G\to H$ be a hom. s.t. 1) $\phi(h)=h\forall h\in H$, 2) $\text{Ker }\phi=K$. Show $G=H\oplus K$

Let $x\in G$. Since $\phi:G\to H$, know that $\phi(x)\in H$ so we can think of $\phi$ as extracting an $H$-content of $x$. Then $\phi(\phi(x))=\phi(x)$ because $\phi$ is identity on $H$. This is the ...

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